Chapter 5. Slope Stability Analyses and Stabilization Measures
- A natural slope is 45-m high and the slope angle is 38 degrees. The surficial soil is loose silty sand with cohesion 45 kN/m2, internal friction angle 25 degrees, and bulk unit weight 18.0 kN/m3. The thickness of the topsoil is 1.0 m in the vertical direction. The slope is dry. Determine the factor of safety of the surficial soil layer against translational failure.
Solution:
Solution using allowable stress design approach:
Since the slope is 45 m high, a think layer of loose silty sand (1.0 m in height) exists on the slope surface, the infinite slope method for dry slope is used. The factor of safety is:
= 5.74
Solution using limit state design approach:
It should be verified that Ed ≤ Rd where Ed is the design effect of the actions (e.g. sliding force) and Rd the design strength.
Following Equation (5.20) and assuming G = 1.35 (Note that this value may change according to local design approach)
kN/m2
Similarly from Equation (5.19) and assuming that R = 1.10, c´ = 1.00, = 1.00 and ´ = 1.00.
45.6 kPa
Since Ed ≤ Rd the limit state is satisfied and the slope is safe.
- A saturated natural slope is 45-m high and the slope angle is 38 degrees. The surficial soil is loose silty sand with effective cohesion 45 kN/m2, effective internal friction angle 25 degrees, and saturated unit weight 19.0 kN/m3. The thickness of the topsoil is 1.0 m in the vertical direction. The downward seepage is parallel to the slope surface. Determine the factor of safety of the surficial soil layer against translational failure.
Solution:
Solution using allowable stress design approach:
The infinite slope method with seepage parallel to the slope face is used. The FS is based on Equation (5.23).
where: c = 45 kN/m2, = 25, sat = 19 kN/m3, = 38, = 199.81= 9.19 kN/m3, H = 1.0 m.
5.17 > 1.5, no slope failure.
Solution using limit state design approach:
Following Equation (5.31) and assuming that G = 1.35,R = 1.10, c´ = 1.00, = 1.00 and ´ = 1.00. Note that these partial factors of safety may change locally.
kPa
kPa
Since Ed ≤ Rd the limit state is satisfied and the slope is safe.
- A reservoir is 45-m deep and the side slope of the reservoir developed a loose surficial layer that is 1.0 m thick (in the vertical direction). The slope angle is 38 degrees. The surficial soil layer has cohesion 45 kN/m2, internal friction angle 25 degrees, and saturated unit weight 19.0 kN/m3. Assuming the reservoir water level is at the top of the side slope, determine the factor of safety of the surficial soil layer against translational failure.
Solution:
Solution using allowable stress design approach:
The infinite slope method with submerged slope without seepage is used.
where: c = 45 kN/m2, = 25, sat = 19 kN/m3, = 38, = 199.81= 9.19 kN/m3, H = 1.0 m.
10.87 > 1.5, no slope failure.
Solution using limit state design approach:
It should be verified that Ed ≤ Rd. Following Equation (5.26), assuming G = 1.35 (Note that this value may change according to local design approach) and recalling that the submerged unit weight has to be used:
kN/m2
43.3 kPa
Since Ed ≤ Rd the limit state is satisfied and the slope is safe.
- A homogeneous earth embankment is 25-meter high and the slope inclination is 2:1 (H:V). The effective cohesion of the embankment is 45 kN/m2, its effective friction angle is 25 degrees, and the bulk unit weight is 19.2 kN/m3. Assume the potential failure surface is a plane. Determine the minimum factor of safety of the slope.
Solution:
Given: H = 25 m, = 26.6, = 19.2 kN/m3, c = 45 kN/m2, = 25.
Since the potential failure surface is a plan, use Culmann’s method and follow the approach in Figure 5.6. The effective stress method is used. The following table is developed for the trial-and-error approach.
,,
Culmann’s method, trial-and-error approach to find FS
FS (assumed) / m / cm (kN/m2) / FSc1.0 / 25 / 0.115 / 390.31
2 / 13.1 / 7.576 / 5.94
3 / 8.8 / 12.933 / 3.48
3.2 / 8.3 / 13.711 / 3.28
3.24 / 8.2 / 13.876 / 3.24
The factor of safety of the slope is 3.24; the slope is stable.
- For the same slope described in Problem 4, what is the factor of safety on a planar failure surface that has an inclination angle of 20 degrees?
Solution:
Given: H = 25 m, = 26.6, = 19.2 kN/m3, c = 45 kN/m2, = 25, and = 20.
Use Culmann’s method and Equation (5.36).
= 3.41
It can be seen that the FS (3.41) on a plane of 20 is larger than the minimum FS of 3.24 that was obtained in Problem 4. Therefore, the plane of 20 inclination is not the critical failure plane.
- A homogeneous earth embankment is to be constructed. The slope inclination is 2:1 (H:V). The effective cohesion of the embankment is 45 kN/m2, its effective friction angle is 25 degrees, and the bulk unit weight is 19.2 kN/m3. Assume the potential failure surface is a plane. If a factor of safety of 1.5 is required, determine the height of the embankment that satisfies the factor of safety.
Solution:
Given: = 26.6, = 19.2 kN/m3, c = 45 kN/m2, = 25, and=1.5
Use Equation (5.45), the height of slope is:
= 203 m
- A saturated and undrained clayey slope is 10-m high and the slope angle is 40 degrees. Subsurface investigation found that the subsoil is homogeneous clay with undrained cohesion of 110 kN/m2 and saturated unit weight of 19.5 kN/m3. A stiff soil layer exists 5 meters below the toe of the slope. A potential toe circle with radius of 20 meters passes the coordinate of (25m, 10m). The toe is at the origin (0, 0).
(1) Determine the factor of safety along the assumed slip circle using the analytical mass method.
(2) Determine the minimum factor of safety of the slope using Taylor’s chart.
Solution:
Given: undrained clayey slope, Hc = 10 m, = 40, = 19.5 kN/m3, cu = 110 kN/m2, assume = 0
(1)
Use AutoCAD, the circular failure surface is drawn, using the two points on the circle and R = 20m. The location of center of the circle, centroid and area (A) of the failure portion can be found. Then, find = 85 = 1.48 rad, and l = 5.88 m, A = 161.67 m2. The AutoCAD drawing is shown above.
Use Equation (5.49), the factor of safety along the potential slip circle using the analytical mass method is:
5.49
(2) The depth of a stiff soil layer from the top of the slope is: ndHc = 10+5=15 m, and Hc = 10 m, so nd = 1.5.
From the Taylor’s chart in Figure 5.9, use nd = 1.5 and = 40 and use interpolation to find Ns 5.85.
From Equation (5.42), find the mobilized undrained shear strength is:
Then:= 3.3 >1.5
The slope is stable.
- A natural slope is shown in Figure 5.41. It contains two soil strata. The slope configuration and the soil characteristics are shown in the figure. A potential toe circle with radius of 70.0 m passes the coordinate of (65m, 30m). The toe is at the origin (0, 0).
(1) Determine the average cohesion, friction angle, and unit weight of the slope and foundation soils.
(2) Determine the factor of safety of the slope along the assumed failure circle.
(3) Determine the minimum factor of safety of the slope, using Taylor’s chart.
Figure 5.41 Slope profile of Problem 8
Solution: not provided.
- A homogeneous silty sand embankment is built on soft clay. The soil profile is shown in Figure 5.42. A potential critical slip surface is also shown.
(1) Use the ordinary method of slices to determine the FS on the assumed failure plane.
(2) Use Bishop’s simplified method of slices to determine the FS on the assumed failure plane, and compare the FS with the one obtained in (1).
(3) Calculate the average cohesion, friction angle, and unit weight of the slope and foundation soil.
(4) Use Taylor’s chart to determine the minimum FS of the slope using the parameters obtained in (3).
Figure 5.42 Slope configuration of Problem 9
Solution:
(1) The ordinary method of slices is used.
The factor of safety is:
The following figure shows the assumed failure circle. The slope is divided into 11 slices. For easy calculation, the bottom of each slice should be entirely in one soil stratum. The widths of the slices can vary. AutoCAD is used to obtain the area, angle i, width, and centroid of each slice. The following table lists the parameters of the slices that are used in the calculation of the FS.
Calculations using the Ordinary Method of Slices
(1) / (2) / (3) / (4) / (5) / (6) / (7) / (8) / (9)Slice number / Slice area in Layer 1 / Slice area in Layer 2 / Total
area / Slice Weight / αi / bi / Wi sinαi /
(m2) / (m2) / (m2) / (kN/m) / (deg) / (m) / (kN/m) / (kN/m)
1 / 124.67 / 0 / 124.67 / 2306.33 / 56.9 / 11.37 / 1932.05 / 1056.84
2 / 254.07 / 0 / 254.07 / 4700.27 / 46.2 / 10 / 3392.47 / 2729.81
3 / 343.98 / 0 / 343.98 / 6363.66 / 36.9 / 10 / 3820.87 / 4270.11
4 / 366.70 / 0 / 366.70 / 6784.04 / 28.6 / 10 / 3247.47 / 4997.91
5 / 124.57 / 0 / 124.57 / 2304.49 / 23.2 / 3.63 / 907.83 / 1777.33
6 / 196.81 / 7.47 / 204.28 / 3782.85 / 19.4 / 6.37 / 1256.52 / 405.28
7 / 240.23 / 35.44 / 275.67 / 5117.52 / 13.5 / 10 / 1194.66 / 617.05
8 / 156.32 / 52.63 / 208.94 / 3891.75 / 6.3 / 10 / 427.06 / 603.65
9 / 77.93 / 76.57 / 154.50 / 2896.61 / -1.4 / 13.29 / -70.77 / 797.45
10 / 0 / 43.18 / 43.18 / 820.44 / -10.4 / 10 / -148.11 / 610.02
11 / 0 / 17.51 / 17.51 / 332.77 / -16.9 / 10 / -96.74 / 627.08
Σ / 15896.32 / 18492.52
1.17
(2) Use Bishop’s simplified method of slices.
The factor of safety is:
The same failure surface and slices are used as in solution (1). Using trial-and-error and an assumed FS of 2.42, the calculations in are prepared in the following table. It is noted that columns (1) to (8) are the same as in Table in solution (1) (ordinary method of slices).
Calculations using Bishop’s simplified method of slices
(1) / (2) / (3) / (4) / (5) / (6) / (7) / (8) / (9) / (10)Slice # / Slice area in Layer 1 / Slice area in Layer 2 / Total
area / Slice Weight / αi / bi / Wi sinαi / mi /
(m2) / (m2) / (m2) / (kN/m) / (deg) / (m) / (kN/m) / (kN/m)
1 / 124.67 / 0 / 124.67 / 2306.33 / 56.9 / 11.37 / 1932.05 / 0.836 / 2111.90
2 / 254.07 / 0 / 254.07 / 4700.27 / 46.2 / 10 / 3392.47 / 0.942 / 3820.68
3 / 343.98 / 0 / 343.98 / 6363.66 / 36.9 / 10 / 3820.87 / 1.008 / 4836.77
4 / 366.70 / 0 / 366.70 / 6784.04 / 28.6 / 10 / 3247.46 / 1.044 / 4978.03
5 / 124.57 / 124.57 / 2304.49 / 23.2 / 3.63 / 907.83 / 1.056 / 1672.15
6 / 196.81 / 7.47 / 204.28 / 3782.85 / 19.4 / 6.37 / 1256.52 / 0.943 / 4415.84
7 / 240.23 / 35.44 / 275.67 / 5117.52 / 13.5 / 10 / 1194.66 / 0.972 / 5879.99
8 / 156.32 / 52.63 / 208.94 / 3891.75 / 6.3 / 10 / 427.06 / 0.994 / 4519.04
9 / 77.93 / 76.57 / 154.50 / 2896.61 / -1.4 / 13.29 / -70.77 / 1.000 / 3694.93
10 / 0 / 43.18 / 43.18 / 820.44 / -10.4 / 10 / -148.11 / 0.984 / 1444.17
11 / 0 / 17.51 / 17.51 / 332.77 / -16.9 / 10 / -96.74 / 0.957 / 974.87
Σ / 15863.32 / 38348.37
2.42 (equal to the assumed FS)
The FS (2.42) obtained using Bishop’s simplified method of slices is larger than the FS (1.17) using the ordinary method of slices.
(3) From the method of slices charge, find:
1 = 75.8 and 2 = 14.2.
Also, c1 = 0, 1 = 40; c2 = 60 kN/m2, 2 = 0.
The average unit weight uses layer thickness as a weighting factor:
Note: the thickness of layer 2 “within” the failure portion can be measured using AutoCAD and is 5.78 m.
(4) Given: Hc = 45 m, = 40, = 18.56 kN/m3, c = 9.47 kN/m2, = 33.7.
The following table is developed for the iterative calculation.
Taylor’s chart for c , use trial-and-error approach to find FS
m (assumed) / /(from Figure 5.10) / (kN/m2) /
10 / 3.78 / 10.2 / 81.9 / 0.11
15 / 2.49 / 15.6
(use extrapolation / 53.5 / 0.176
25 / 1.43 / 21
(use extrapolation) / 39.8 / 0.23
The above table shows that the FS and FSc do not converge. In fact, when a slope overlies a weak foundation of clay with = 0, Duncan and Wright (2005) suggested the above averaging procedures shall not be used. With a foundation soil of = 0, the critical circle usually extends below the toe into the foundation. If using the above procedure, a non-zero would be produced for the foundation soil, resulting in a different and erroneous location of the critical failure surface. To resolve this, Duncan and Wright (2005) suggested the shear stress on an approximate slip surface in each layer be first calculated using:
where: is the average normal stress on the slip surface in layer i.
For layer 1:
For layer 2:
i is treated as the undrained shear strength su(i) for the layer. Then, Equation (5.69) can be used to obtain the equivalent average undrained shear strength for the entire slope and foundation:
Then, the slope and foundation are taken as a homogeneous undrained slope and Taylor’s chart (Fig 5.9) for undrained clay with = 0 is used to obtain FS.
The assumed failure surface extends 5.78 m below the toe, so:
(use 1.2)
= 40
Find Ns 6.0.
From Equation (5.42),
Then:= 2.18 > 1.5
It is noted that the obtained FS is between the factors of safety using the Fellenius method and the Bishop simplified method of slices.
- A homogeneous natural slope is 20 m high. The same type of soil extends to great depth. The slope angle is 35 degrees. The effective cohesion is 60 kN/m2, the effective internal friction angle is 25 degrees. The bulk density is 1800 kg/m3. Use Bishop’s simplified method of slices to determine the minimum factor of safety, using:
(1) One slope circle.
(2) One toe circle.
(3) One deep-seated, mid-point circle.
Solution: Not provided.
- A natural slope is 35 meters high and the slope inclination is 2(H):1(V). The slope’s properties are: c = 50 kN/m2, = 25, and = 19.0 kN/m3. No firm layer was encountered beneath the slope during the subsurface investigation. Downward steady seepage that is parallel to the slope surface occurs in the slope. The average pore water pressure ratio for the slope was determined to be 0.4.
(1) Use Bishop-Morgenstern method to determine the minimum factor of safety (FS) for the slope.
(2) Use Spencer charts to determine the FS.
(3) Use Michalowski charts to determine the FS.
Solution:
Given: H = 35 m, slope inclination 2:1 and find slope angle = 26.6, = 19.0 kN/m3, c = 50 kN/m2, = 25, ru = 0.4.
(1) Use Bishop-Morgenstern method and consider the effect of pore pressure.
, it is between 0.025 and 0.05.
Given the slope inclination of 2:1, = 25, and average pore pressure ratio ru = 0.4,
the following trial-and-error approach is used to find the minimum FS:
- From Table 5.6(d): assume a failure surface passes D = 1.00:
m = 1.872, n = 1.386. 1.318
- From Table 5.6(e): assume a failure surface passes D = 1.25:
m = 2.004, n = 1.641.
- From Table 5.6(f): assume a failure surface passes D = 1.50
m = 2.308, n = 1.914. 1.542
Therefore, the minimum FS = 1.31 and the bottom of the critical failure surface is at the same level as the toe (D = 1.0).
(2) Use Spencer chart and consider the effect of pore pressure
ru = 0.4 is between 0.25 and 0.5. The factor of safety for ru = 0.25 and ru = 0.5 are determined first. Then linear interpolation is used to determine the FS for ru = 0.4. Using the approach in Figure 5.19, the following tables are developed.
Determination of FS for ru = 0.25
Fassume / m using / (from Figure 5.18b) / Fcalculate1.5 / 17.3 / 0.052 / 1.44
1.48 / 17.5 / 0.05 / 1.50
1.49 / 17.4 / 0.051 / 1.47
Choose FS = 1.48
Determination of FS for ru = 0.5
Fassume / m using / (from Figure 5.18c) / Fcalculate1.5 / 17.3 / 0.082 / 0.91
1.2 / 21.2 / 0.069 / 1.09
1.15 / 22.0 / 0.066 / 1.14
Choose FS = 1.14
Use linear interpolation, FS at ru = 0.4:
1.27
(3) Use Michalowski charts
ru = 0.4 is between 0.25 and 0.5. The factor of safety for ru = 0.25 and ru = 0.5 are determined first. Then linear interpolation is used to determine the FS for ru = 0.4.
, = 26.6
From Figure 5.20(d), using linear interpolation, find: =3.5.
So: 1.63 for ru = 0.25.
From Figure 5.20(f), using linear interpolation, find: =2.5.
So: 1.16 for ru = 0.5.
Use linear interpolation, FS at ru = 0.4:
1.34
- A sandy soil embankment is 10 meters high and the slope inclination is 2(H):1(V). The slope’s properties are: c = 0 kN/m2, = 30, and = 19.5 kN/m3. A firm layer exists 5 meters beneath the toe of the slope. The average pore water pressure ratio for the slope is 0.25.
(1) Use Bishop-Morgenstern method to determine the minimum factor of safety (FS) for the slope.
(2) Use Spencer charts to determine the FS.
(3) Use Michalowski charts to determine the FS.
Solution:
Given: H = 10 m, slope inclination 2:1 and find slope angle = 26.6, = 19.5 kN/m3, c = 0, = 30, ru = 0.25.
(1) Use Bishop-Morgenstern method and consider the effect of pore pressure.
From Table 5.6(a), for all values of D (that is, for all potential failure surfaces), find m = 1.155, n = 1.444. m and n are for the critical failure surface.
The minimum 0.57 <1.0. The slope will fail.
(2) Use Spencer chart and consider the effect of pore pressure
Given: ru = 0.25, , and slope inclination = 2:1, from Figure 5.18(b), find required m = 35.
So: 0.82 < 1.0. The slope will fail.
(3) Use Michalowski charts
Given: ru = 0.25 , , = 26.6
From Figure 5.20(d), using linear interpolation, find: =1.7.
So: 0.98 < 1.0. The slope will fail.
- A rapid drawdown is needed for a reservoir. The side slope for the reservoir is 84- meter high, its cohesion is 62 kN/m2, the internal friction angle is 25 degrees, and the saturated unit weight is 18.5 kN/m3. The inclination of the slope is 3:1 (H:V). Assume the initial water level is at the top of the side slope; a 42-meter drawdown is needed. Determine the factor of safety of the side slope under the rapid drawdown condition.
Solution:
Morgenstern charts for rapid drawdown are used.
is between 0.025 and 0.05. FS for and are first determined, then linear interpolation is used to find FS for .
Drawdown ratio:
Use Figure 5.23(b), for and given = 25, find FS = 1.3.
Use Figure 5.24(b), for and given = 25, find FS = 1.6.
For 1.42 < 1.5
- A rapid drawdown is needed for a reservoir. The side slope for the reservoir is 80- meter high and its inclination is 2:1. The slope’s properties are: c = 40 kN/m2, = 30, and = 20 kN/m3. The initial water level is at the top of the side slope. In order to maintain a factor of safety of 1.5 for the side slope, what is the maximum drawdown depth (L)?
Solution:
Morgenstern charts for rapid drawdown are used.
Use Figure 5.23(a), and given = 30 and FS = 1.5, find drawdown ratio:
So, the maximum drawdown depth L = 0.2 80 = 16 m